Respuesta :

shadrachadamu

Answer:

∫101/(x+1)dx=(1−0)∫101/(x+1)dx/(1−0)=∫101/(x+1)f(x)dx=E(1/(x+1))

Where f(x)=1, 0

And then I calculated log 2 from the calculator and got 0.6931471806

From R, I got 0.6920717

So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.

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